

A354531


Numbers k such that 2*(2^k1) is in A354525.


4



1, 2, 3, 5, 7, 9, 13, 17, 19, 31, 61, 67, 89, 107, 127, 137
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OFFSET

1,2


COMMENTS

Numbers k such that for every prime factor p of 2^k1 we have gpf(2*(2^k1)+p) = p.
Numbers k such that for every prime factor p of 2^k1, 2*(2^k1)+p is psmooth.
All terms except 2 are odd: if k is even, then 3 is a factor of 2^k1, so 3^m = 2*(2^k1)+3 = 2^(k+1) + 1 => k+1 >= 3^(m1). The only possible case is (k,m) = (2,2).
Clearly A000043 is a subsequence. The exceptional terms (1, 9, 67, 137, ...) are listed in A354532.
The next term is >= 349. The next composite term, if exists, is >= 7921 = 89^2.


LINKS

Table of n, a(n) for n=1..16.


EXAMPLE

See A354532.


PROG

(PARI) gpf(n) = vecmax(factor(n)[, 1]);
ispsmooth(n, p, {lim=1<<256}) = if(n<=lim, n==1  gpf(n)<=p, my(N=n/p^valuation(n, p)); forprime(q=2, p, N=N/q^valuation(N, q); if((N<=lim && isprime(N))  N==1, return(N<=p))); 0); \\ check if n is psmooth, using brute force if n is too large
isA354531(n, {lim=256}, {p_lim=1<<32}) = {
my(N=2^n1);
if(isprime(N), return(1));
if(n>lim, forprime(p=3, p_lim, if(N%p==0 && !ispsmooth(2*N+p, p), return(0)))); \\ first check if there is a prime factor p <= p_lim of 2^n1 such that 2*(2^n1)+p is not psmooth (for large n)
my(d=divisors(n));
for(i=1, #d, my(f=factor(2^d[i]1)[, 1]); for(j=1, #f, if(!ispsmooth(2*N+f[j], f[j], 1<<lim), return(0)))); 1 \\ then check if 2*(2^n1)+p is psmooth for p2^d1, dN
}


CROSSREFS

Cf. A354525, A000043, A354532, A354533, A354536.
Sequence in context: A028870 A338356 A057886 * A302835 A200672 A341497
Adjacent sequences: A354528 A354529 A354530 * A354532 A354533 A354534


KEYWORD

nonn,hard,more


AUTHOR

Jianing Song, Aug 16 2022


STATUS

approved



